Write the augmented matrix of the system and use the matrix method to solve the system. $$ \begin{array}{rr} x+y+z= & 210 \ x-2 y= & 0 \ 2 x+3 y+5 z= & 570 \ 2 x-y+z= & 210 \end{array} $$ Select the correct choice below and, if necessary, fill in the answer boxes to complete your choice. A. This system has exactly one solution. The solution is $$ \square $$ , $$ \square $$ , $$ ) $$ ). (Type an exact answer in simplified form.) B. This system has infinitely many solutions of the form $$ \square $$ , ,z), where $$ z $$ is any real number. (Type expressions using $$ z $$ as the variable.) C. This system has no solution.

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To solve the given system of equations using the matrix method, we first need to write the augmented matrix for the system. The system of equations is:

$$
\begin{array}{rcl}
x + y + z & = & 210 \
x - 2y & = & 0 \
2x + 3y + 5z & = & 570 \
2x - y + z & = & 210
\end{array}
$$

### Step 1: Write the Augmented Matrix

The augmented matrix for this system can be written as follows:

$$
\begin{bmatrix}
1 & 1 & 1 & | & 210 \
1 & -2 & 0 & | & 0 \
2 & 3 & 5 & | & 570 \
2 & -1 & 1 & | & 210
\end{bmatrix}
$$

### Step 2: Row Reduction

Next, we will perform row operations to reduce this matrix to row echelon form.

1. Subtract Row 1 from Row 2:
   $$
   R_2 = R_2 - R_1 \implies R_2 = [0, -3, -1 | -210]
   $$

2. Subtract 2 times Row 1 from Row 3:
   $$
   R_3 = R_3 - 2R_1 \implies R_3 = [0, 1, 3 | 150]
   $$

3. Subtract 2 times Row 1 from Row 4:
   $$
   R_4 = R_4 - 2R_1 \implies R_4 = [0, -3, -1 | -210]
   $$

Now the matrix looks like this:
$$
\begin{bmatrix}
1 & 1 & 1 & | & 210 \
0 & -3 & -1 & | & -210 \
0 & 1 & 3 & | & 150 \
0 & -3 & -1 & | & -210
\end{bmatrix}
$$

4. Now, we can simplify Row 2 and Row 4:
   $$
   R_2 = \frac{1}{-3}R_2 \implies R_2 = [0, 1, \frac{1}{3} | 70]
   $$
   $$
   R_4 = R_4 - R_2 \implies R_4 = [0, 0, 0 | 0]
   $$

Now the matrix looks like this:
$$
\begin{bmatrix}
1 & 1 & 1 & | & 210 \
0 & 1 & \frac{1}{3} & | & 70 \
0 & 0 & 0 & | & 0
\end{bmatrix}
$$

5. Substitute Row 2 into Row 1:
   $$
   R_1 = R_1 - R_2 \implies R_1 = [1, 0, \frac{2}{3} | 140]
   $$

Now the matrix looks like this:
$$
\begin{bmatrix}
1 & 0 & \frac{2}{3} & | & 140 \
0 & 1 & \frac{1}{3} & | & 70 \
0 & 0 & 0 & | & 0
\end{bmatrix}
$$

### Step 3: Back Substitution

From the reduced matrix, we can express the variables in terms of $$ z $$:

1. From Row 1: 
   $$
   x + \frac{2}{3}z = 140 \implies x = 140 - \frac{2}{3}z
   $$

2. From Row 2:
   $$
   y + \frac{1}{3}z = 70 \implies y = 70 - \frac{1}{3}z
   $$

### Conclusion

The system has infinitely many solutions of the form:
$$
(x, y, z) = \left(140 - \frac{2}{3}z, 70 - \frac{1}{3}z, z\right)
$$

Thus, the correct choice is:
**B. This system has infinitely many solutions of the form $$ (140 - \frac{2}{3}z, 70 - \frac{1}{3}z, z) $$, where $$ z $$ is any real number.**
The system has infinitely many solutions of the form $$ (140 - \frac{2}{3}z, 70 - \frac{1}{3}z, z) $$, where $$ z $$ is any real number.

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